A ship is sailing due west at \(V\) knots while a plane, with an airspeed of \(kV\) knots, where \(k>\sqrt{2},\) patrols so that it is always to the north west of the ship. If the wind in the area is blowing from north to south at \(V\) knots and the pilot is instructed to return to the ship every thirty minutes, how long will her outward flight last? Assume that the maximum distance of the plane from the ship during the above patrol was \(d_{w}\) miles. If the air now becomes dead calm, and the pilot's orders are maintained, show that the ratio \(d_{w}/d_{c}\) of \(d_{w}\) to the new maximum distance, \(d_{c}\) miles, of the plane from the ship is \[ \frac{k^{2}-2}{2k(k^{2}-1)}\sqrt{4k^{2}-2}. \]
The random variables \(X\) and \(Y\) are independently normally distributed with means 0 and variances 1. Show that the joint probability density function for \((X,Y)\) is \[ \mathrm{f}(x,y)=\frac{1}{2\pi}\mathrm{e}^{-\frac{1}{2}(x^{2}+y^{2})}\qquad-\infty < x < \infty,-\infty < y < \infty. \] If \((x,y)\) are the coordinates, referred to rectangular axes, of a point in the plane, explain what is meant by saying that this density is radially symmetrical. The random variables \(U\) and \(V\) have a joint probability density function which is radially symmetrical (in the above sense). By considering the straight line with equation \(U=kV,\) or otherwise, show that \[ \mathrm{P}\left(\frac{U}{V} < k\right)=2\mathrm{P}(U < kV,V > 0). \] Hence, or otherwise, show that the probability density function of \(U/V\) is \[ \mathrm{g}(k)=\frac{1}{\pi(1+k^{2})}\qquad-\infty < k < \infty. \]
A message of \(10^{k}\) binary digits is sent along a fibre optic cable with high probabilities \(p_{0}\) and \(p_{1}\) that the digits 0 and 1, respectively, are received correctly. If the probability of a digit in the original message being a 1 is \(\alpha,\) find the probability that the entire message is received correctly. Find the probability \(\beta\) that a randomly chosen digit in the message is received as a 1 and show that \(\beta=\alpha\) if, and only if \[ \alpha=\frac{q_{0}}{q_{1}+q_{0}}, \] where \(q_{0}=1-p_{0}\) and \(q_{1}=1-p_{1}.\) If this condition is satisfied and the received message consists entirely of zeros, what is the probability that it is correct? If now \(q_{0}=q_{1}=q\) and \(\alpha=\frac{1}{2},\) find the approximate value of \(q\) which will ensure that a message of one million binary digits has a fifty-fifty chance of being received entirely correctly. The probability of error \(q\) is proportional to the square of the length of the cable. Initially the length is such that the probability of a message of one million binary bits, among which 0 and 1 are equally likely, being received correctly is \(\frac{1}{2}.\) What would this probability become if a booster station were installed at its mid-point, assuming that the booster station re-transmits the received version of the message, and assuming that terms of order \(q^{2}\) may be ignored?
A candidate finishes examination questions in time \(T\), where \(T\) has probability density function \[ \mathrm{f}(t)=t\mathrm{e}^{-t}\qquad t\geqslant0, \] the probabilities for the various questions being independent. Find the moment generating function of \(T\) and hence find the moment generating function for the total time \(U\) taken to finish two such questions. Show that the probability density function for \(U\) is \[ \mathrm{g}(u)=\frac{1}{6}u^{3}\mathrm{e}^{-u}\qquad u\geqslant0. \] Find the probability density function for the total time taken to answer \(n\) such questions.
Solution: \begin{align*} && M_T(x) &= \mathbb{E}[e^{xT}] \\ &&&= \int_0^{\infty} e^{xt}te^{-t} \d t \\ &&&= \int_0^{\infty}te^{(x-1)t} \d t \\ &&&= \left [ \frac{t}{x-1} e^{(x-1)t} \right]_0^{\infty} - \int_0^\infty \frac{e^{(x-1)t}}{x-1} \d t \\ &&&= \left [ \frac{e^{(x-1)t}}{(x-1)^2} \right]_0^{\infty} \\ &&&= \frac{1}{(x-1)^2} \\ \\ && M_U(x) &= M_{T_1+T_2}(x) \\ &&&= \frac1{(x-1)^4} \\ \\ && I_n &= \int_0^{\infty} t^ne^{(x-1)t} \d t \\ &&&= \left[ \frac{1}{(x-1)}t^ne^{(x-1)t} \right]_0^{\infty} - \frac{n}{(x-1)} \int_0^{\infty}t^{n-1}e^{(x-1)t} \d t \\ &&&= -\frac{n}{(x-1)}I_{n-1} \\ \Rightarrow && I_n &= \frac{n!}{(1-x)^{n+1}} \\ \\ \Rightarrow && \int_0^{\infty} e^{xt} \frac16u^3e^{-u} \d u &= \int_0^{\infty} \frac16u^3e^{(x-1)u} \d u \\ &&&= \frac{1}{(1-x)^4} \\ \Rightarrow && f_U(u) &= \frac16u^3e^{-u} \\ \\ && M_{X_1+\cdots+X_n}(x) &= \frac{1}{(x-1)^{2n}} \\ \Rightarrow && f_{X_1+\cdots+X_n}(t) &= \frac1{(2n-1)!} t^{2n-1}e^{-t} \end{align*} (NB: This is the gamma distribution)